If (1,3) is the midpoint of a chord of the ci | vedclass

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(B) The equation of the circle is $x^2+y^2-4x-8y+16=0$.Given the midpoint of the chord is $(x_1, y_1) = (1,3)$.The equation of the chord with midpoint

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$8$

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(B) The equation of the circle is $x^2+y^2-4x-8y+16=0$.Given the midpoint of the chord is $(x_1, y_1) = (1,3)$.The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$ and $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.Substituting the values: $x(1) + y(3) - 2(x+1) - 4(y+3) + 16 = 1^2 + 3^2 - 4(1) - 8(3) + 16$.$x + 3y - 2x - 2 - 4y - 12 + 16 = 1 + 9 - 4 - 24 + 16$.$-x - y + 2 = -2$.$-x - y = -4$,which simplifies to $x + y = 4$.The chord intersects the coordinate axes at $A(0,4)$ and $B(4,0)$.The triangle formed by the chord and the coordinate axes is a right-angled triangle with base $OB = 4$ and height $OA = 4$.$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4 = 8$.

If $(1,3)$ is the midpoint of a chord of the circle $x^2+y^2-4x-8y+16=0$,then the area of the triangle formed by that chord with the coordinate axes is

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